Let x0 of type ι → ο be given.
Let x1 of type ι → ι → ι be given.
Let x2 of type ι → ι → ι be given.
Assume H0: ∀ x3 x4 . x0 x3 ⟶ x0 x4 ⟶ x0 (x1 x3 x4).
Assume H1: ∀ x3 x4 x5 . x0 x3 ⟶ x0 x4 ⟶ x0 x5 ⟶ x2 x3 (x1 x4 x5) = x1 (x2 x3 x4) (x2 x3 x5).
Assume H2: ∀ x3 x4 x5 . x0 x3 ⟶ x0 x4 ⟶ x0 x5 ⟶ x2 (x1 x3 x4) x5 = x1 (x2 x3 x5) (x2 x4 x5).
Let x3 of type ι be given.
Let x4 of type ι be given.
Let x5 of type ι be given.
Let x6 of type ι be given.
Let x7 of type ι be given.
Let x8 of type ι be given.
Let x9 of type ι be given.
Assume H3: x0 x3.
Assume H4: x0 x4.
Assume H5: x0 x5.
Assume H6: x0 x6.
Assume H7: x0 x7.
Assume H8: x0 x8.
Assume H9: x0 x9.
Apply unknownprop_55de5c79fadd89ca3e161a61e8ef1cc68aeee5eba6c4fec4d11d6eacbce11bf5 with
x0,
x1,
x2,
x3,
x4,
x5,
x6,
x7,
x1 x8 x9,
λ x10 x11 . x11 = x1 (x1 (x2 x3 x8) (x2 x3 x9)) (x1 (x1 (x2 x4 x8) (x2 x4 x9)) (x1 (x1 (x2 x5 x8) (x2 x5 x9)) (x1 (x1 (x2 x6 x8) (x2 x6 x9)) (x1 (x2 x7 x8) (x2 x7 x9))))) leaving 9 subgoals.
The subproof is completed by applying H0.
The subproof is completed by applying H2.
The subproof is completed by applying H3.
The subproof is completed by applying H4.
The subproof is completed by applying H5.
The subproof is completed by applying H6.
The subproof is completed by applying H7.
Apply H0 with
x8,
x9 leaving 2 subgoals.
The subproof is completed by applying H8.
The subproof is completed by applying H9.
Apply H1 with
x3,
x8,
x9,
λ x10 x11 . x1 x11 (x1 (x2 x4 (x1 x8 x9)) (x1 (x2 x5 (x1 x8 x9)) (x1 (x2 x6 (x1 x8 x9)) (x2 x7 (x1 x8 x9))))) = x1 (x1 (x2 x3 x8) (x2 x3 x9)) (x1 (x1 (x2 x4 x8) (x2 x4 x9)) (x1 (x1 (x2 x5 x8) (x2 x5 x9)) (x1 (x1 (x2 x6 x8) (x2 x6 x9)) (x1 (x2 x7 x8) (x2 x7 x9))))) leaving 4 subgoals.
The subproof is completed by applying H3.
The subproof is completed by applying H8.
The subproof is completed by applying H9.
Apply H1 with
x4,
x8,
x9,
λ x10 x11 . x1 (x1 (x2 x3 x8) (x2 x3 x9)) (x1 x11 (x1 (x2 x5 (x1 x8 x9)) (x1 (x2 x6 (x1 x8 x9)) (x2 x7 (x1 x8 x9))))) = x1 (x1 (x2 x3 x8) (x2 x3 x9)) (x1 (x1 (x2 x4 x8) (x2 x4 x9)) (x1 (x1 (x2 x5 x8) (x2 x5 x9)) (x1 (x1 ... ...) ...))) leaving 4 subgoals.